Question

In an experiment to estimate the acceleration due to gravity, a student drops a ball at a distance of 1 m above the floor. His lab partner measures the time it takes to fall and obtains an estimate of 0.46 s.

(a) What do they estimate the acceleration due to gravity to be?
(b) Let s be the ball's position relative to the floor. Using the value of the acceleration due to gravity that they obtained, and assuming that the ball is released at t = 0. determines (in m) as a function of time.

Answer 1

Answer:

$9.45179\ m/s^2$

$s=4.725895t^2$

Explanation:

t = Time taken = 0.46

u = Initial velocity

v = Final velocity

s = Displacement = 1 m

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0\times 0.46+\frac{1}{2}\times a\times 0.46^2\\\Rightarrow a=\frac{1\times 2}{0.46^2}\\\Rightarrow a=9.45179\ m/s^2$

The acceleration due to gravity is 9.45179 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=\frac{1}{2}at^2\\\Rightarrow s=\dfrac{1}{2}9.45179t^2\\\Rightarrow s=4.725895t^2$

The function is $s=4.725895t^2$

Answer 2

(a) The estimated acceleration due to gravity is 9.45 m/s²

(b) the position of the ball as a function of time is given by $s=4.275t^2$

Equation of motion:

Given that the ball takes 0.46s to drop from 1 m height.

(a) From Second equation of motion:

$s=ut+\frac{1}{2}at^2$

where s is the distance,

a is the acceleration,

and t is the time taken

the initial velocity u=0, so:

$1=\frac{1}{2}\times a\times(0.46)^2\\\\a=9.45m/s^2$is the estimated acceleration due to gravity.

(b) the ball's position as a function of time can be given as:

$s=ut+\frac{1}{2} at^2\\\\s=0.5\times9.45t^2\\\\s=4.725t^2$

Learn more about equation of motion:

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